1. ## Arrangements question

Question:

A boy has five blue marbles, four green marbles and three red marbles. In how many ways can be arrange four of them in a row, if the marbles of any one colour are indistinguishable?

The if blue 1...... blue 5 are all indistinguishable this is the same for green and red

thus the answer would be $\displaystyle \frac{12!}{5!4!3!} = 27720$ but the answer is 80 in the book where did i go wrong?

2. ## Re: Arrangements question Originally Posted by bigmansouf Question:
A boy has five blue marbles, four green marbles and three red marbles. In how many ways can be arrange four of them in a row, if the marbles of any one colour are indistinguishable?
The if blue 1...... blue 5 are all indistinguishable this is the same for green and red
thus the answer would be $\displaystyle \frac{12!}{5!4!3!} = 27720$ but the answer is 80 in the book where did i go wrong?
You need to rewrite the question. I think that it means that just four of the marbles are in the row.
All blue, all green, three blue & one green, three blue & one red etc.

3. ## Re: Arrangements question

4B 1 way

3B 8 ways

2B 24 ways

1B 32 ways

no blue 15 ways

4. ## Re: Arrangements question Originally Posted by Idea 4B 1 way

3B 8 ways

2B 24 ways

1B 32 ways

no blue 15 ways

My way of thinking

5 Blue 4 Green 3 Red

All blue B B B B - 1 way

3 blues B B B
for the last place We can choose Green 4 times or Red 3 times, I thought the the answer will be 7 since there are 7 (4 green + 3 red) for a person to choose from
how is it 8 ways

2 blues B B
for the last two places We can choose Green 4 times or Red 3 times or a combination of both

so B B G R
4 3 = 4 X 3 = 12 ways
or B B G G
4 3 = 4 X 3 = 12 ways
or B B R R
4 3 = 4 X 3 = 12 ways

12 + 12 + 12 = 36 ways

I dont understand how you arrive at each answer

5. ## Re: Arrangements question

These are the 8 ways with 3 blue marbles

B B B G
B B G B
B G B B
G B B B

B B B R
B B R B
B R B B
R B B B

note that the marbles of any one colour are indistinguishable